An infinite tower of powers

Idea. The tower contains a copy of itself as its exponent, so its value $T$ satisfies $x^T=T$. Solve for $x$, then make sure the sequence actually converges to that value.

Step 1: the equation. With $T=2$, $x^2=2$, so $x=\sqrt{2}$.

Step 2: convergence. For $x=\sqrt{2}$, let $a_1=\sqrt{2}$ and $a_{n+1}={\sqrt{2}}^{a_n}$. If $a_n<2$, then $a_{n+1}<{\sqrt{2}}^2=2$, and $a_{n+1}>a_n$ as long as $a_n<2$ (the function ${\sqrt{2}}^a-a$ is positive on $[0,2)$). So the sequence increases and stays below 2; its limit $L$ satisfies ${\sqrt{2}}^L=L$, and the only solution in $(0,2]$ is $L=2$.

So $x=\sqrt{2}\approx 1.414$.

The trap. The same algebra for $T=4$ gives $x^4=4$, so $x=\sqrt{2}$ again. But the tower of $\sqrt{2}$ converges to 2, not 4: the equation $x^T=T$ is necessary, not sufficient. In fact the tower converges only for $e^{-e}\le x\le e^{1/e}$, and its value is then at most $e$, so no tower equals 4.